I got a novelty cube at as swag at a convention once; it was like a normal cube, but with slogans written on the squares of the cube. If you solved the cube, you could read the slogan.

This got me to wondering if it's possible to solve this cube such that some of the words faced the wrong way.

If you have a standard Rubik's cube, and draw arrows on each label (all arrows on each face point in the same direction), is it possible to solve the cube such that some of the arrows face different directions from the original orientation? If so, how many distinct solved cubes are there?

Note: obviously, for all the non-center faces, there is only one possible orientation. For example, if the red-blue edge has "up" on the blue face pointing at red, then the red-blue-yellow and red-blue-green corners must be next to that edge, and must also have "up" on the blue faces point toward red. Thus, the orientations of the edges and corners is fixed, but it might be possible for the centers to end up rotated.

2 Answers
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If you have a standard Rubik's cube, and draw arrows on each label (all arrows on each face point in the same direction), is it possible to solve the cube such that some of the arrows face different directions from the original orientation?

Yes, this is possible. The centres can be rotated.
To rotate the U-face centre by 180 degrees, you can do the move sequence RL U2 R'L' U twice. To rotate the U-face centre a clockwise quarter turn and the F-face centre an anti-clockwise quarter turn, do FB' LR' UD' F' U'D L'R F'B U. A simpler but much longer sequence for this is to repeat U F' 63 times.

If so, how many distinct solved cubes are there?

Each of the 6 centres has 4 possible orientations. However, not all $4^6=4096$ orientations are achievable. It is not possible to turn a single centre 90 degrees without permuting the corner and edge pieces. In fact, the total amount of twist applied to the centres must be a multiple of 180 degrees, i.e. an even number of quarter turns. This is because an odd number of quarter turns of the centres would require an odd number of quarter turn moves of the faces, and that would apply an odd permutation to the corners and to the edges, so they cannot return to their initial locations.

So the centres only have at most $\frac{4^6}{2}=2048$ achievable states. With the move sequences mentioned above, every such state can be solved, so all those $2048$ states can indeed be achieved.